Question 331650
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If I take what you wrote literally, then I suspect we are going to come out very wrong here.  You see *[tex \Large X] and *[tex \Large x] are two very different things.  Hence, in


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ g(x)\ =\ X^2\ -\ 4x]


*[tex \LARGE X^2] is just treated as a constant because you are defining the function *[tex \LARGE g] as a function of lower case *[tex \LARGE x] as the independent variable.


Therefore


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ g(2)\ =\ X^2\ -\ 4(2)\ =\ X^2\ -\ 8]


but I very much doubt that is what you meant.


On the other hand, if you really meant:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ g(x)\ =\ x^2\ -\ 4x]


Then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ g(2)\ =\ (2)^2\ -\ 4(2)\ =\ 4\ -\ 8\ =\ -4]


By the way, instead of typing out "squared," use the caret mark (^) to indicate raising to a power.  Your function would have been properly rendered g(x) = x^2 - 4x


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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